Pretest

Solution

• Let G be a graph. Prove or disprove:
  • Deleting a vertex of maximum degree cannot raise the average degree.
    True. We prove it directly. Let the number of vertices be n, the sum of all vertex degrees be S, and the maximum degree be m (which is at most n-1); the average degree is then S/n. Now remove a vertex of degree m (maximum degree); we obtain a graph of n-1 vertices, whose degree sum is S-2m (we subtract m twice, since each of the m edges incident upon the deleted vertex contributed to the degree of both of its endpoints) and whose average degree is thus (S-2m)/(n-1). Now we have
    n*(n-1)* [ S/n - (S-2m)/(n-1) ] = (n-1)*S - n*(S-2m) = 2mn-S
    But this quantity cannot be negative and can only be 0 when all vertices have degree m. Hence the new average degree is no larger than the old.
  • Deleting a vertex of minimum degree cannot lower the average degree.
    False. As a simple counterexample, consider the graph formed by two vertices joined by an edge (formally, this graph is K₂). Both vertices have minimum degree (degree 1) and the average degree is 1. Remove one vertex: the result is a graph with a single vertex (K₁) -- that vertex has degree 0 and that is the average degree, which has thus been lowered.
• Let G be a connected graph and let k be the maximum length of a simple path in G. Prove that, if simple paths P1 and P2 both have length k, then they have at least one vertex in common.
  Assume that the two paths do not have any vertex in common. Since the graph is connected, it is possible to move from one path to the other; since the two paths are vertex-disjoint, there is at least one edge in the connection between the two that does not belong to either path. Say the connection is between vertex u on the first path and vertex v on the second, along a third (disjoint) path from u to v. Vertex u partitions the first path into two pieces, so choose the larger (if they have the same number of edges, choose arbitrarily); ditto for vertex v. Now consider moving from the end of the longer piece from the first path to u, then to v, then to the end of the longer piece from the second path: the resulting trajectory contains at least one more edge than either of the given paths, and so we have found a simple path of length longer than k, a contradiction.
• Let G be a graph having no cycle of even length. Prove that each edge of G appears in at most one cycle.
  Another proof by contradiction. Suppose edge e appears in cycles C1 and C2. Consider moving from one endpoint of e along C1 until the other endpoint of e is reached; instead of now crossing e and returning to the start (as C1 would do), move along C2 instead and follow it until it returns to the start vertex. We have described a new cycle; its length is the sum of the lengths of C1 and C2, minus 2 (because e never gets crossed to complete either cycle). But l(C1)+l(C2) is the sum of two odd numbers and so is even; and subtracting 2 from it keeps it even, so our new cycle has even length, the desired contradiction.
• Consider the following game. A prize is placed behind one of three (closed) doors. The player chooses a door, which remains closed for now; the game master then opens one of the other two doors -- always choosing one that does not contain a prize (if the player chose a wrong door, the game master opens the one remaining wrong door; if the player chose the right door, the game master chooses one of the other two doors uniformly at random). The game master then tells the player "As you can see, there is no prize behind this door I just opened; now do you want to stay with your original choice or switch to the other unopened door?" Decide what is the right strategy and prove it by calculating the odds.
  The key to the problem is to realize that, if the player picked the wrong door at the beginning, then switching guarantees getting the prize! (Conversely, if the player picked the right door, then clearly switching guarantees losing.)\ \ The player has a 1/3 chance of picking the correct door; if the player refuses to switch, 1/3 is then the odds of winning. If the player always switches however, the situation is better: with probability 1/3 (of having picked the correct door in the first place), the player will lose, but with probability 2/3 (of having picked a wrong door), the player will win, so the odds of winning with a switch are 2/3, twice as high as for staying with the original choice.
• From n equally spaced points on a circle (in real coordinates), three distinct points are chosen uniformly at random. What is the probability that they form an equilateral triangle? an isosceles triangle? a triangle with sides of all different lengths?
  Since the points are equally spaced, we can only get an equilateral triangle when the number of points is a multiple of 3, so let us assume n=3k for some k. Note that there are (n choose 3) possible triangles (roughly n³/6). But there are very equilateral ones: once we have chosen one vertex of the triangle, the other two are forced; moreover, choosing any of the three that form such a triangle forces the same triple, so there really are only n/3 equilateral triangles. Thus the probability of getting an equilateral triangle is roughly 2/n². Isosceles triangles can be obtained by picking one of the two equal sides along with the summit of the triangle, at which point the rest is forced. We have n choices for the summit, then roughly n/2 choices for the other endpoint (the other n/2 choices are symmetric) -- it is rough because n/2 may not be integer and also because it ignores the equilateral triangles with their higher symmetry, but, as we have seen, the equilateral triangles are too few to affect the asymptotic values. Thus we have roughly n²/2 isosceles triangles, so that the probability of getting one is roughly 3/n -- much larger than for equilateral triangles, but still vanishingly small. Finally, the probability of getting a triangle with sides of all different lengths is just (1-p_iso), or roughly (n-3)/n, which converges rapidly to 1. It is not too hard to get exact formulas, but not really worth it.
• Prove that every infinite subset of an infinite countable set is countable -- i.e., prove that, is set S can be placed in a 1-1 correspondence (bijection) with the natural integers, then so can any infinite subset of S.
  Because S is in 1-1 correspondence with N, we might as well call it N. S' has a natural injection into S: the identity (map each element of S' to itself, now regarded as an element of S). Unfortunately, this map is not surjective and thus not a bijection; this is what we need to fix. We can define the new bijection recursively; in words, it just renumbers the elements of S', a subset of S=N, starting from 1 and consecutively. Formally, define f: S' -> N with
    f(i) = f(j)+1, where j = max k {k < i and k belongs to S'}
• Prove that every positive integer can be written as a sum of all different Fibonacci numbers.
  First let us look at a brute-force proof. We prove this by induction. 1 and 2 are Fibonacci numbers by themselves and a fortiori meet the requirements; assume then that all numbers up to and including n can be written as a sum of distinct Fibonacci numbers. That's not quite enough: we need a more careful look at what that expression can be. We have a standard way of reducing (simplifying) a sum of Fibonacci numbers: the sum of two consecutive ones is just the next one up -- the recursive definition of Fibonacci numbers. This means that we can assume that the sum in question never includes two consecutive Fibonacci numbers. For instance, we could write 3 = F₂+F₁, but that is just F₃; similarly, we have 10 = F₄+F₃+F₂, but we write the simpler 10 = F₅+F₂. Let E(F,n) be the expression denoting the sum for the value n -- and recall that it does not include consecutive terms. Then we write
  n+1 = E(F,n) + 1 = E(F,n) + F₀
  since F₀=1. The expression E(F,n)+F₀ might already be a valid E(F,n+1) -- when E(F,n) included neither F₀ nor F₁ (it could not have included both, of course). Otherwise we start a process of replacing consecutive Fibonacci numbers by the next one up; this process obviously terminates when we reach the largest Fibonacci number in E(F,n) and it is well defined, since we cannot generate a second copy of a Fibonacci number in E(F,n), thanks to the fact that E(F,n) does not contain consecutive Fibonacci numbers.
  Now for a more elegant proof, which actually proves the same additional result (no consecutive numbers in the sum). The proof is also by induction, with the same basis. Now if given value n+1, either it is a Fibonacci number, say F_i, in which case we are done, or it is not, in which case, if the largest Fibonacci number less than n+1 is F_j, we can write n+1 as the sum F_j + (sum for (n+1-F_j). The second sum exists by inductive hypothesis and cannot contain the jth Fibonacci number, so the resulting sum obeys our conditions. This completes the proof; to verify that the sum can always avoid consecutive Fibonacci numbers, it is enough to see that, in the decomposition F_j + (sum for (n+1-F_j), if the sum for (n+1-F_j) contained F_j-1, then n+1 would be larger than F_j+F_j-1=F_j+1, but we assumed that F_j was the largest Fibonacci number less than n+1.
• You are given a stack of cards numbered 1 through n; the stack is randomly ordered. Play the following game: if the top card has value i, reverse the order of the top i cards on the stack; continue until the top card has value 1, at which point the game stops. Prove that the game always stops, no matter what the initial order was, and prove that, if k distinct cards show up at the top during the process, then the game stops in at most F_k-1 steps, where F_n is the nth Fibonacci number.
  The hardest question -- or, more precisely, the only nontrivial question. Note that the second proof subsumes the first: it simply gives a numerical upper bound on the number of moves that can be made before stopping and thus of course implies that the games always stops. Thus it suffices to prove the second, more detailed result. We prove it by induction on k, the number of distinct cards that appear at the top of the pack (the pack has n cards). The base case, with k=1, is trivially correct -- there cannot be any reversal, so the top card is already 1. Assume then that the claim holds for up to and including k-1 distinct cards appearing on top; call these cards c₁, c₂, c₃, ..., c_k-1, in increasing order of values (of course, that is not directly related to order of appearance on top, not to mention that the same card can appear many times on top). Note that we have c₁=1 -- because we know that the game ended and so that card 1 ended up on top; moreover, that card appeared on top only once, of course, at the very end. Thus a perhaps more productive way to view the inductive hypothesis is that a game in which at most k-1 different cards appear on top can proceed for at most F_k-1-2 moves without reaching the final state. Now consider a game in which exactly one more card shows on top and call these cards c₁, c₂, c₃, ..., c_k, in increasing order of values (they need not be the same values as our previous c_i). By induction, if only the first k-1 of these cards showed on top, the game would stop and stop in at most F_k-1-1 moves; and one of the first k-1 cards would then have value 1. But that card can only show on top at the very end - as soon as it shows, the game stops. So, before card c_k shows on top the first time, only cards c₂, c₃, ..., c_k-1 can show on top; and after card c_k shows on top the first time, only cards c₂, c₃, ..., c_k can show on top, with card c₁ making it to the top in the last, additional move. Now we can write a recurrence to describe the process: when a total of k different cards appear on top during a game, only k-2 different cards can show on top until the first appearance of card c_k, then only k-1 different cards appear on top until the very last move, when card c₁ could move to the top. We have an additional move from the first to the second part, so the recurrence is
  nbr(k) = nbr(k-2) + 1 + nbr(k-1)
  where nbr(k) denotes the number of moves that can be made with k different cards appearing on top without reaching a final state. By induction we have
  nbr(k) = nbr(k-2) + 1 + nbr(k-1) = F_k-2-1 + 1 + F_k-1-1
  and thus
  nbr(k) = F_k-2-1 + 1 + F_k-1-1 = F_k-1
  as desired.